Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the cardinality of a transcendence basis of $\mathbb{C}$ over $\mathbb{Q}$? Is it that of the continuum? Proof?

share|cite|improve this question
@HagenvonEitzen Maybe because I answered my own questions. Or maybe because there are so many unanswered questions. Or maybe because I'm choosy. Or maybe because of all of them. – Makoto Kato Oct 28 '12 at 21:48
Yes. Zorn's lemma. – Andrés E. Caicedo Oct 28 '12 at 21:49
One can show that the algebraic closure of an infinite set $S$ of complex numbers has the same cardinality as $S$. – André Nicolas Oct 28 '12 at 22:00
Please avoid placing the question only in the title. The body of the question should stand on its own. – Asaf Karagila Oct 28 '12 at 23:43
The title is just a one line meant to pitch your question to the people browsing the main page. The body needs to stand on its own accord. – Asaf Karagila Oct 29 '12 at 1:13
up vote 2 down vote accepted

If $S$ has infinite cardinality $\kappa$, then $|\mathbb Q(S)|=\kappa$. And if $|F|=\kappa$, then $|F[X]|=\kappa $ and finally $|\bar F|=\kappa$. Therefore we need exactly $\kappa=2^{\aleph_0}$ if we want $\overline{\mathbb Q(S)}=\mathbb C$.

share|cite|improve this answer

(Assume the axiom of choice)

If $V$ is a vector space over $F$, and $B$ is a basis for $V$, and at least $B$ or $F$ are infinite, then every $v\in V$ is identified with a unique finite function which is a subset of $F\times B$. Namely, the unique decomposition of $v$ to non-zero scalars and basis elements.

We therefore have that $|V|\leq |F\times B|^{<\omega}=|F\times B|=\max\{|F|,|B|\}$, and on the other hand $|F|\leq|V|$ and $|B|\leq|V|$, therefore: $$\max\{|F|,|B|\}\leq|V|\leq\max\{|F|,|B|\}\implies |V|=\max\{|F|,|B|\}$$

Apply this to the case of $\mathbb Q$ and $\mathbb C$, we have that $|\mathbb C|=2^{\aleph_0}$ and therefore the basis has to have size $2^{\aleph_0}$.

Without assuming the axiom of choice it is consistent that there is no such basis, so we cannot talk too much about that. It is also open whether the existence of a basis implies the well-orderability of $\mathbb C$. If it does, then the above theorem holds and it has to have size continuum; if it doesn't then it's hard to say too much, although it is likely to be provable that its size is again $2^{\aleph_0}$.

share|cite|improve this answer
Please note that a transcendence basis of a field extension is not a vector space basis. – Makoto Kato Oct 29 '12 at 15:11
@Makoto: Yes. You are correct of course. The same argument applies in this case, though. You need to slightly modify it to discuss algebraic independence rather than linear independence. However since for infinite fields the ring of polynomials has the same cardinality as the field this the argument above applies. I have to leave now, but later tonight I will correct this post. – Asaf Karagila Oct 29 '12 at 15:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.